Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $r = \dfrac{t^2 - 13t + 36}{-3t^2 + 6t + 189} \times \dfrac{4t + 28}{t + 3} $
Explanation: First factor out any common factors. $r = \dfrac{t^2 - 13t + 36}{-3(t^2 - 2t - 63)} \times \dfrac{4(t + 7)}{t + 3} $ Then factor the quadratic expressions. $r = \dfrac {(t - 9)(t - 4)} {-3(t - 9)(t + 7)} \times \dfrac {4(t + 7)} {t + 3} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac { (t - 9)(t - 4) \times 4(t + 7)} { -3(t - 9)(t + 7) \times (t + 3)} $ $r = \dfrac {4(t - 9)(t - 4)(t + 7)} {-3(t - 9)(t + 7)(t + 3)} $ Notice that $(t - 9)$ and $(t + 7)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {4\cancel{(t - 9)}(t - 4)(t + 7)} {-3\cancel{(t - 9)}(t + 7)(t + 3)} $ We are dividing by $t - 9$ , so $t - 9 \neq 0$ Therefore, $t \neq 9$ $r = \dfrac {4\cancel{(t - 9)}(t - 4)\cancel{(t + 7)}} {-3\cancel{(t - 9)}\cancel{(t + 7)}(t + 3)} $ We are dividing by $t + 7$ , so $t + 7 \neq 0$ Therefore, $t \neq -7$ $r = \dfrac {4(t - 4)} {-3(t + 3)} $ $ r = \dfrac{-4(t - 4)}{3(t + 3)}; t \neq 9; t \neq -7 $